3x squared minus 6x plus 11=0

The quadratic formula is:

$$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$

Your equation only has complex roots:

$${\mathtt{3}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{6}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{11}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{6}}}}{\mathtt{\,\times\,}}{i}{\mathtt{\,-\,}}{\mathtt{3}}\right)}{{\mathtt{3}}}}\\
{\mathtt{x}} = {\frac{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{6}}}}{\mathtt{\,\times\,}}{i}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}\right)}{{\mathtt{3}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}\left({\mathtt{\,-\,}}{\mathtt{0.999\: \!999\: \!999\: \!999\: \!821}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1.632\: \!993\: \!161\: \!854\: \!606\: \!3}}{i}\right)\\
{\mathtt{x}} = {\mathtt{0.999\: \!999\: \!999\: \!999\: \!821}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1.632\: \!993\: \!161\: \!854\: \!606\: \!3}}{i}\\
\end{array} \right\}$$

The calculator here has introduced some rounding error; the 0.99999.... parts should be exactly 1

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No real solutions...